GAV-sound with conjunctive queries

Transcription

1 GAV-sound with conjunctive queries Source and global schema as before: source R 1 (A, B),R 2 (B,C) Global schema: T 1 (A, C), T 2 (B,C) GAV mappings become sound: T 1 {x, y, z R 1 (x,y) R 2 (y,z)} T 2 R 2 Let D exact be the unique database that arises from the exact setting (with replaced by =) Then every database D sound that satisfies the sound setting also satisfies D exact D sound L. Libkin 1 Data Integration and Exchange

2 GAV-sound with conjunctive queries cont d Conjunctive queries are monotone: D 1 D 2 Q(D 1 ) Q(D 2 ) Exact solution is a sound solution too, and is contained in every sound solution. Hence certain answers for each conjunctive query certain(d,q) = Q(D sound ) = Q(D exact ) D sound The solution for GAV-exact gives us certain answers for GAV-sound, for conjunctive (and more generally, monotone) queries. L. Libkin 2 Data Integration and Exchange

3 Query answering using views General setting: database relations R 1,...,R n. Several views V 1,...,V k are defined as results of queries over the R i s. We have a query Q over R 1,...,R n. Question: Can Q be answered in terms of the views? In other words, can Q be reformulated so it only refers to the data in V 1,...,V k? L. Libkin 3 Data Integration and Exchange

4 Query answering using views in data integration LAV: R 1,...,R n are global schema relations; Q is the global schema query V i s are the sources defined over the global schema We must answer Q based on the sources (virtual integration) GAV: R 1,...,R n are the sources that are not fully available. Q is a query in terms of the source (or a query that was reformulated in terms of the sources) Must see if it is answerable from the available views V 1,...,V k. We know the problem is impossible to solve for full relational algebra, hence we concentrate on conjunctive queries. L. Libkin 4 Data Integration and Exchange

5 Conjunctive queries: rule-based notation We often write conjunctive queries as logical statements: {t,y, r d ( Movie(t, d,y) RV(t, r) y > 2000 ) } Rule-based: Q(t,y, r) : Movie(t, d,y), RV(t,r),y > 2000 Q(t,y, r) is the head of the rule Movie(t,d, y), RV(t, r),y > 2000 is its body conjunctions are replaced by commas variables that occur in the body but not in the head (d) are assumed to be existentially quantified essentially logic programming notation (without functions) L. Libkin 5 Data Integration and Exchange

6 Query answering using views: example Two relations in the database: Cites(A,B) (if A cites B), and SameTopic(A,B) (if A, B work on the same topic) Query Q(x, y) : SameTopic(x, y), Cites(x, y), Cites(y, x) Two views are given: V 1 (x, y) : Cites(x,y), Cites(y,x) V 2 (x, y) : SameTopic(x,y), Cites(x, x ), Cites(y, y ) Suggested rewriting: Q (x,y) : V 1 (x,y), V 2 (x,y) Why? Unfold using the definitions: Q (x, y) : Cites(x,y), Cites(y,x), SameTopic(x,y), Cites(x,x ), Cites(y, y ) Equivalent to Q L. Libkin 6 Data Integration and Exchange

7 Query answering using views Need a formal technique (algorithm): cannot rely on the semantics. Query Q: SELECT R1.A FROM R R1, R R2, S S1, S S2 WHERE R1.A=R2.A AND S1.A=S2.A AND R1.A=S1.A AND R1.B=1 and S2.B=1 Q(x) : R(x, y), R(x, 1),S(x,z), S(x, 1) Equivalent to Q(x) : R(x, 1), S(x, 1) So if we have a view V (x,y) : R(x, y),s(x, y) (i.e. V = R S), then Q = π A (σ B=1 (V )) Q can be rewritten (as a conjunctive query) in terms of V L. Libkin 7 Data Integration and Exchange

8 Query rewriting Setting: Queries V 1,...,V k over the same schema σ (assume to be conjunctive; they define the views) Each Q i is of arity n i A schema ω with relations of arities n 1,...,n k Given: a query Q over σ a query Q over ω Q is a rewriting of Q if for every σ-database D, Q(D) = Q ( V 1 (D),...,V k (D) ) L. Libkin 8 Data Integration and Exchange

9 Maximal rewriting Sometimes exact rewritings cannot be obtained Q is a maximally-contained rewriting if: it is contained in Q: Q ( V 1 (D),...,V k (D) ) Q(D) for all D it is maximal such: if Q ( V 1 (D),...,V k (D) ) Q(D) for all D, then Q Q L. Libkin 9 Data Integration and Exchange

10 Side remark: query rewriting and certain answers If we have sources R = (R 1,...,R k ), we can view conditions V 1 (D) = R 1,..., V k (D) = R k as an incomplete specification of a database D To answer Q over D, given R 1,...,R k, we want to compute certain answers: certain(q,r) = { Q(D) V1 (D) = R 1,..., V k (D) = R k } If for every such D we have Q(D) = Q (V 1 (D),...,V k (D)), then certain(q,r) = Q. But we may even look at a more general way of query answering by finding a rewriting Q so that certain(q,r) = Q (R) L. Libkin 10 Data Integration and Exchange

11 Query rewriting: a naive algorithm Given: conjunctive queries V 1,...,V k over schema σ a query Q over σ Algorithm: guess a query Q over the views Unfold Q in terms of the views Check if the unfolding is contained in Q If one unfolding is equivalent to Q, then Q is a rewriting Otherwise take the union of all unfoldings contained in Q it is a maximally contained rewriting L. Libkin 11 Data Integration and Exchange

12 Why is it not an algorithm yet? Problem 1: we do not yet know how to test containment and equivalence. But we shall learn soon Problem 2: the guess stage. There are infinitely many conjunctive queries. We cannot check them all. Solution: we only need to check a few. L. Libkin 12 Data Integration and Exchange

13 Guessing rewritings A basic fact: If there is a rewriting of Q using V 1,...,V k, then there is a rewriting with no more conjuncts than in Q. E.g., if Q(x) : R(x, y),r(x, 1),S(x,z), S(x, 1), we only need to check conjunctive queries over V with at most 4 conjuncts. Moreover, maximally contained rewriting is obtained as the union of all conjunctive rewritings of length of Q or less. Complexity: enumerate all candidates (exponentially many); for each an NP (or exponential) algorithm. Hence exponential time is required. Cannot lower this due to NP-completeness. L. Libkin 13 Data Integration and Exchange

14 Containment and optimization of conjunctive queries Reminder: conjunctive queries = SPJ queries = rule-based queries = simple SELECT-FROM-WHERE SQL queries (only AND and equality in the WHERE clause) Extremely common, and thus special optimization techniques have been developed Reminder: for two relational algebra expressions e 1, e 2, e 1 = e 2 is undecidable. But for conjunctive queries, even e 1 e 2 is decidable. Main goal of optimizing conjunctive queries: reduce the number of joins. L. Libkin 14 Data Integration and Exchange

15 Optimization of conjunctive queries: an example Given a relation R with two attributes A, B Two SQL queries: Q1 Q2 SELECT R1.B, R1.A FROM R R1, R R2 WHERE R2.A=R1.B SELECT R3.A, R1.A FROM R R1, R R2, R R3 WHERE R1.B=R2.B AND R2.B=R3.A Are they equivalent? If they are, we saved one join operation. In relational algebra: Q 1 = π 2,1 (σ 2=3 (R R)) Q 2 = π 5,1 (σ 2=4 4=5 (R R R)) L. Libkin 15 Data Integration and Exchange

16 Optimization of conjunctive queries cont d Are Q 1 and Q 2 equivalent? If they are, we cannot show it by using equivalences for relational algebra expression. Because: they don t decrease the number of or operators, but Q 1 has 1 join, and Q 2 has 2. But Q 1 and Q 2 are equivalent. How can we show this? But representing queries as databases. This representation is very close to rule-based queries. Q 1 (x,y) : R(y, x),r(x, z) Q 2 (x,y) : R(y, x),r(w, x),r(x, u) L. Libkin 16 Data Integration and Exchange

17 Conjunctive queries into tableaux Tableau: representing of a conjunctive query as a database We first consider queries over a single relation Q 1 (x,y) : R(y, x),r(x, z) Q 2 (x,y) : R(y, x),r(w, x),r(x, u) Tableaux: A B y x x z x y answer line A B y x w x x u x y answer line Variables in the answer line are called distinguished L. Libkin 17 Data Integration and Exchange

18 Tableau homomorphisms A homomorphism of two tableaux f : T 1 T 2 is a mapping f : {variables of T 1 } {variables of T 2 } {constants} For every distinguished x, f(x) = x For every row x 1,...,x k in T 1, f(x 1 ),...,f(x k ) is a row of T 2 Query containment: Q Q if Q(D) Q (D) for every database D Homomorphism Theorem: Let Q,Q be two conjunctive queries, and T,T their tableaux. Then Q Q if and only if there exists a homomorphism f : T T L. Libkin 18 Data Integration and Exchange

19 Applying the Homomorphism Theorem: Q 1 = Q 2 T1 A B y x x z x y T2 A B y x w x x u x y f(x)=x, f(y)=y f(u)=z, f(w)=y Hence Q1 Q2 T1 A B y x x z x y T2 A B y x w x x u x y f(x)=x, f(y)=y f(z)=u Hence Q2 Q1 L. Libkin 19 Data Integration and Exchange

20 Applying the Homomorphism Theorem: Complexity Given two conjunctive queries, how hard is it to test if Q 1 = Q 2? it is easy to transform them into tableaux, from either SPJ relational algebra queries, or SQL queries, or rule-based queries But testing the existence of a homomorphism between two tableaux is hard: NP-complete. Thus, a polynomial algorithm is unlikely to exists. However, queries are small, and conjunctive query optimization is possible in practice. L. Libkin 20 Data Integration and Exchange

21 Minimizing conjunctive queries Goal: given a conjunctive query Q, find an equivalent conjunctive query Q with the minimum number of joins. Assume Q is Q( x) : R 1 ( u 1 ),...,R k ( u k ) Assume that there is an equivalent conjunctive query Q of the form with l < k Q ( x) : S 1 ( v 1 ),...,S l ( v l ) Then Q is equivalent to a query of the form Q ( x) : R i1 ( u i1 ),...,R l ( u il ) In other words, to minimize a conjunctive query, one has to delete some subqueries on the right of : L. Libkin 21 Data Integration and Exchange

22 Minimizing conjunctive queries cont d Given a conjunctive query Q, transform it into a tableau T Let Q be a minimal conjunctive query equivalent to Q. Then its tableau T is a subset of T. Minimization algorithm: T := T repeat until no change choose a row t in T if there is a homomorphism f : T T {t} then T := T {t} end Note: if there exists a homomorphism T T {t}, then the queries defined by T and T {t} are equivalent. Because: there is always a homomorphism from T {t} to T. (Why?) L. Libkin 22 Data Integration and Exchange

23 Minimizing SPJ/conjunctive queries: example R with three attributes A, B,C SPJ query Q = π AB (σ B=4 (R)) π BC (π AB (R) π AC (σ B=4 (R))) Equivalently, a SQL query: SELECT R1.A, R2.B, R3.C FROM R R1, R R2, R R3 WHERE R1.B=4 AND R2.A=R3.A AND R3.B=4 AND R2.B=R1.B Translate into a conjunctive query: Rule-based: x 1, z 1,z 2 (R(x, 4,z 1 ) R(x 1, 4, z 2 ) R(x 1, 4,z) y = 4) Q(x,y, z) : R(x, 4,z 1 ), R(x 1, 4,z 2 ), R(x 1, 4, z),y = 4 L. Libkin 23 Data Integration and Exchange

24 Minimizing SPJ/conjunctive queries cont d Tableau T: A B C x 4 z 1 x 1 4 z 2 x 1 4 z x 4 z Minimization, step 1: is there a homomorphism from T to A B C x 1 4 z 2 x 1 4 z x 4 z Answer: No. For any homomorphism f, f(x) = x (why?), thus the image of the first row is not in the small tableau. L. Libkin 24 Data Integration and Exchange

25 Minimizing SPJ/conjunctive queries cont d Step 2: Is T equivalent to A B C x 4 z 1 4 z x 4 z x 1 Answer: Yes. Homomorphism f: f(z 2 ) = z, all other variables stay the same. The new tableau is not equivalent to A B C A B C x 4 z 1 or x 1 4 z x 4 z x 4 z Because f(x) = x,f(z) = z, and the image of one of the rows is not present. L. Libkin 25 Data Integration and Exchange

26 Minimizing SPJ/conjunctive queries cont d Minimal tableau: A B C x 4 z 1 4 z x 4 z x 1 Back to conjunctive query: An SPJ query: Q (x, y, z) : R(x, y, z 1 ), R(x 1, y,z), y = 4 SELECT R1.A, R1.B, R2.C FROM R R1, R R2 WHERE R1.B=R2.B AND R1.B=4 π AB (σ B=4 (R)) π BC (σ B=4 (R)) L. Libkin 26 Data Integration and Exchange

27 Review of the journey We started with π AB (σ B=4 (R)) π BC (π AB (R) π AC (σ B=4 (R))) Translated into a conjunctive query Built a tableau and minimized it Translated back into conjunctive query and SPJ query Applied algebraic equivalences and obtained π AB (σ B=4 (R)) π BC (σ B=4 (R)) Savings: one join. L. Libkin 27 Data Integration and Exchange

28 All minimizations are equivalent Let Q be a conjunctive query, and Q 1, Q 2 two conjunctive queries equivalent to Q Assume that Q 1 and Q 2 are both minimal, and let T 1 and T 2 be their tableaux. Then T 1 and T 2 are isomorphic; that is, T 2 can be obtained from T 1 by renaming of variables. That is, all minimizations are equivalent. In particular, in the minimization algorithm, the order in which rows are considered, is irrelevant. L. Libkin 28 Data Integration and Exchange

29 Equivalence of conjunctive queries: the general case So far we assumed that there is only one relation R, but what if there are many? Construct tableaux as before: Tableau: B: A B x y x Q(x,y): B(x,y), R(y, z), R(y, w),r(w, y) y R: A B y z y w w y Formally, a tableau is just a database where variables can appear in tuples, plus a set of distinguished variables. L. Libkin 29 Data Integration and Exchange

30 Tableaux and multiple relations Given two tableaux T 1 and T 2 over the same set of relations, and the same distinguished variables, a homomorphism h : T 1 T 2 is a mapping f : {variables of T 1 } {variables of T 2 } such that - f(x) = x for every distinguished variable, and - for each row t in R in T 1, f( t) is in R in T 2. Homomorphism theorem: let Q 1 and Q 2 be conjunctive queries, and T 1, T 2 their tableaux. Then Q 2 Q 1 if and only if there exists a homomorphism f : T 1 T 2 L. Libkin 30 Data Integration and Exchange

31 Minimization with multiple relations The algorithm is the same as before, but one has to try rows in different relations. Consider homomorphism f(z) = w, and f is the identity for other variables. Applying this to the tableau for Q yields B: A B x y x y R: A B y w w y This cannot be further reduced, as for any homomorphism f, f(x) = x, f(y) = y. Thus Q is equivalent to One join is eliminated. Q (x, y) : B(x,y), R(y, w),r(w, y) L. Libkin 31 Data Integration and Exchange

32 Query rewriting Recall the algorithm, for a given Q and view definitions V 1,...,V k : Look at all rewritings that have as at most as many joins as Q check if they are contained in Q take the union of those that are This is the maximally contained rewriting There are algorithms that prune the search space and make looking for rewritings contained in Q more efficient the bucket algorithm MiniCon May see of them later L. Libkin 32 Data Integration and Exchange

33 How hard is it to answer queries using views? Setting: we now have an actual content of the views. As before, a query is Q posed against D, but must be answered using information in the views. Suppose I 1,...,I k are view instances. Two possibilities: Exact mappings: I j = V j (D) Sound mappings: I j V j (D) We need certain answers for given I = (I 1,...,I k ): certain exact (Q, I) = Q(D) certain sound (Q, I) = D: I j =V j (D) for all j D: I j V j (D) for all j Q(D) L. Libkin 33 Data Integration and Exchange

34 How hard is it to answer queries using views? If certain exact (Q, I) or certain sound (Q, I) are impossible to obtain, we want maximally contained rewritings: Q (I) certain exact (Q, I), and if Q (I) certain exact (Q, I) then Q (I) Q (I) (and likewise for sound) How hard is it to compute this from I? In databases, we reason about complexity in two ways: The big-o notation (O(n log n) vs O(n 2 ) vs O(2 n )) Complexity-theoretic notions: PTIME, NP, DLOGSPACE, etc Advantage of complexity-theoretic notions: if you have a O(2 n ) algorithm, is it because the problem is inherently hard, or because we are not smart enough to come up with a better algorithm (or both)? L. Libkin 34 Data Integration and Exchange

35 Complexity classes: what you always wanted to know but never dared to ask Or a 5/5-introduction: a five minute review that tells you what are likely to remember 5 years after taking a complexity theory course. The big divide: PTIME (computable in polynomial time, i.e. O(n k ) for some fixed k) Inside PTIME: tractable queries (although high-degree polynomial are intractable) Outside PTIME: intractable queries (efficient algorithms are unlikely) Way outside PTIME: provably intractable queries (efficient algorithms do not exist) EXPTIME: c n -algorithms for a constant c. Could still be ok for not very large inputs Even further 2-EXPTIME: c cn. Cannot be ok even for small inputs (compare 2 10 and ). L. Libkin 35 Data Integration and Exchange

36 Inside PTIME AC 0 TC 0 NC 1 DLOG NLOG PTIME AC 0 : very efficient parallel algorithms (constant time, simple circuits) relational calculus TC 0 : very efficient parallel algorithms in a more powerful computational model with counting gates basic SQL (relational calculus/grouping/aggregation) NC 1 : efficient parallel algorithms regular languages DLOG: very little O(log n) space is required SQL + (restricted) transitive closure NLOG: O(log n) space is required if nondeterminism is allowed SQL + transitive closure (as in the SQL3 standard) L. Libkin 36 Data Integration and Exchange

37 Beyond PTIME PTIME { NP conp } PSPACE PTIME: can solve a problem in polynomial time NP: can check a given candidate solution in polynomial time another way of looking at it: guess a solution, and then verify if you guessed it right in polynomial time conp: complement of NP verify that all reasonable candidates are solutions to a given problem. Appears to be harder than NP but the precise relationship isn t known PSPACE: can be solved using memory of polynomial size (but perhaps an exponential-time algorithm) L. Libkin 37 Data Integration and Exchange

38 Complete problems These are the hardest problems in a class. If our problem is as hard as a complete problem, it is very unlikely it can be done with lower complexity. For NP: SAT (satisfiability of Boolean formulae) many graph problems (e.g. 3-colourability) Integer linear programming etc For PSPACE: Quantified SAT Two XML DTDs are equivalent L. Libkin 38 Data Integration and Exchange

39 Complexity of query answering We want the complexity of finding in terms of the size of I certain exact (Q, I) or certain sound (Q, I) If all view definitions are conjunctive queries and Q is a relational algebra or a SQL query, then the complexity is conp. (blackboard) This is too high! If all view definitions are conjunctive queries and Q is a conjunctive query, then the complexity is PTIME. Because: the maximally contained rewriting computes certain answers! L. Libkin 39 Data Integration and Exchange

40 Complexity of query answering query language view language CQ CQ relational calculus CQ ptime conp undecidable CQ ptime conp undecidable relational calculus undecidable undecidable undecidable CQ conjunctive queries CQ conjunctive queries with inequalities (for example, Q(x) : R(x, y), S(y, z),x z ) L. Libkin 40 Data Integration and Exchange

41 Complexity of query answering: conp-completeness idea Start with a graph G this is our instance D is G together with a colouring, with 3 colours; each node is assigned one colour. Q asks if we have an edge (a,b) with a b and a,b of the same colour. If G is not 3-colourable, then every instance D would satisfy Q Otherwise, if G is 3-colourable, we can find extensions that are and that are not 3-colourable hence certain answers are empty. Thus if we can compute certain answers, we can test non-3-colourability conp-completeness. L. Libkin 41 Data Integration and Exchange